Question

Nitrogen dioxide reacts with water to produce oxygen and ammonia:4NO2(g)+6H2O(g)→7O2(g)+4NH3(g)

At a temperature of 415 ∘C and a pressure of 725 mmHg , how many grams of NH3 can be produced when 4.10 L of NO2 react?

Glucose, C6H12O6, is metabolized in living systems according to the reaction C6H12O6(s)+6O2(g)→6CO2(g)+6H2O(l)

How many grams of water can be produced from the reaction of 24.5 g of glucose and 6.30 L of O2 at 1.00 atm and 37 ∘C?

Answer 1

The balanced equation is:

4NO2(g)+6H2O(g)→7O2(g)+4NH3(g)

Number of moles of NO2, n = PV/RT, Based on the gas laws, PV = nRT,

R = 0.082 L-atm/(K-mol) and temperature is in Kelvin, pressure is in atm.

n = (725/760) atm *4.10 L/[0.082 (L-atm/K-mol)*(415+273)K]

=n (725/760)*4.10/[0.082*(415+273)] mols

= 0.0693 mols

Now, 4 mols of NO2 produces 4 mols of NH3 gas, so the ratio is 1:1,

Therefore, the number of moles of NH3 will be produced = 0.0693 mols.

Molar mass of NH3 = 17.031 g/mol

Amount of NH3 gas produced = 0.0693mols*17.031g/mol = 1.180 g